NCERT Class 10 Maths Chapter 1: Real Numbers Cheatsheet
Key Concepts & Formulas
- Fundamental Theorem of Arithmetic: Every composite number can be expressed as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.
- HCF (Highest Common Factor): Product of the smallest power of each common prime factor in the numbers.
- LCM (Least Common Multiple): Product of the greatest power of each prime factor involved in the numbers.
- Relation: \( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \) (for two numbers).
- Irrational Numbers Proof: For prime \( p \), \( \sqrt{p} \) is irrational. Assume \( \sqrt{p} = \frac{a}{b} \) (coprime \( a,b \)), leads to contradiction using theorem: if \( p \) divides \( a^2 \), \( p \) divides \( a \).
- Composite Numbers: Numbers that can be expressed as product of primes.
- Ending with Zero: Requires factors 2 and 5 in prime factorization.
Quick Reference Table
| Concept | Formula/Method | Example |
|---|---|---|
| Prime Factorization | Break into product of primes | \( 32760 = 2^3 \times 3^2 \times 5 \times 7 \times 13 \) |
| HCF | Smallest powers of common primes | \( \text{HCF}(6,20) = 2^1 = 2 \) |
| LCM | Greatest powers of all primes | \( \text{LCM}(6,20) = 2^2 \times 3 \times 5 = 60 \) |
| HCF for Three Numbers | Same method extended | \( \text{HCF}(6,72,120) = 2 \times 3 = 6 \) |
| LCM for Three Numbers | Greatest powers | \( \text{LCM}(6,72,120) = 2^3 \times 3^2 \times 5 = 360 \) |
| Irrationality Proof | Contradiction assuming rational | For \( \sqrt{2} \): Assume \( \frac{a}{b} \), \( 2b^2 = a^2 \), 2 divides a and b, contradiction. |
| Composite Check | Product of smaller integers >1 | \( 7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1) = 13 \times 78 \), composite. |
Tips & Advice
Prime Factorization First: Always start with prime factorization for HCF, LCM, or checking divisibility.
Irrational Proofs: Use proof by contradiction and the theorem that if a prime divides a square, it divides the number.
HCF/LCM Relation: Use to find one if the other is known, e.g., \( \text{LCM} = \frac{\text{product}}{\text{HCF}} \).
Ending Digits: For ending with zero, check for 2 and 5 in factors; otherwise impossible.
Verify Solutions: After calculating HCF/LCM, verify product equals numbers' product.
Composite Expressions: Factor out common terms to show compositeness.
Real Numbers Quick Tips
- Unique Factorization: Prime factors are unique except for order.
- HCF × LCM = Product: Only for two numbers; not directly for three or more.
- Irrational Sums/Products: If rational + irrational = rational, leads to contradiction.
- Cyclic Events: Time to meet again is LCM of individual times.
- Prove Irrational: Assume \( \frac{p}{q} \), square, show common factor contradiction.
Real Numbers FAQ
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What is the Fundamental Theorem of Arithmetic?Every composite number can be uniquely factorized into primes, apart from order.
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How to find HCF and LCM using prime factorization?HCF: smallest powers of common primes; LCM: greatest powers of all primes.
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Why is \( \sqrt{2} \) irrational?Proof by contradiction: assuming rational leads to 2 dividing both numerator and denominator infinitely.
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Can a number like 6^n end with digit 0?No, lacks factor 5 in prime factorization.
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What is the relation between HCF and LCM?For two numbers a and b, HCF(a,b) × LCM(a,b) = a × b.
HCF & LCM Calculator
Additional Informatics: Key Theorems and Applications
- Theorem 1.2: If p (prime) divides a^2, then p divides a (a positive integer).
- Applications of FTA: Proving irrationality (e.g., \( \sqrt{2}, \sqrt{3} \)), determining decimal expansions of rationals.
- Euclid's Division Algorithm: Any positive integer a = bq + r, 0 ≤ r < b; used for HCF via repeated division.
- Real Numbers Classification: Rationals (p/q form) and Irrationals (non-repeating, non-terminating decimals).
- History Note: Carl Friedrich Gauss proved the uniqueness in FTA in 1801.
Real Numbers Speed Quiz
Test your knowledge with 5 questions from NCERT Class 10 Maths Chapter 1! You have 30 seconds per question.
Solutions to Exercises
Exercise 1.1
1. Express each number as a product of its prime factors:
(i) 140 = \( 2^2 \times 5 \times 7 \)
(ii) 156 = \( 2^2 \times 3 \times 13 \)
(iii) 3825 = \( 3^2 \times 5^2 \times 17 \)
(iv) 5005 = \( 5 \times 7 \times 11 \times 13 \)
(v) 7429 = \( 17 \times 19 \times 23 \)
2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 = \( 2 \times 13 \), 91 = \( 7 \times 13 \), HCF = 13, LCM = \( 2 \times 7 \times 13 = 182 \), Verification: \( 13 \times 182 = 2366 \), \( 26 \times 91 = 2366 \)
(ii) 510 = \( 2 \times 3 \times 5 \times 17 \), 92 = \( 2^2 \times 23 \), HCF = 2, LCM = \( 2^2 \times 3 \times 5 \times 17 \times 23 = 23460 \), Verification: \( 2 \times 23460 = 46920 \), \( 510 \times 92 = 46920 \)
(iii) 336 = \( 2^4 \times 3 \times 7 \), 54 = \( 2 \times 3^3 \), HCF = \( 2 \times 3 = 6 \), LCM = \( 2^4 \times 3^3 \times 7 = 3024 \), Verification: \( 6 \times 3024 = 18144 \), \( 336 \times 54 = 18144 \)
3. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12 = \( 2^2 \times 3 \), 15 = \( 3 \times 5 \), 21 = \( 3 \times 7 \), HCF = 3, LCM = \( 2^2 \times 3 \times 5 \times 7 = 420 \)
(ii) 17 = 17, 23 = 23, 29 = 29, HCF = 1, LCM = \( 17 \times 23 \times 29 = 11339 \)
(iii) 8 = \( 2^3 \), 9 = \( 3^2 \), 25 = \( 5^2 \), HCF = 1, LCM = \( 2^3 \times 3^2 \times 5^2 = 1800 \)
4. Given that HCF (306, 657) = 9, find LCM (306, 657).
LCM = \( \frac{306 \times 657}{9} = 22338 \)
5. Check whether 6^n can end with the digit 0 for any natural number n.
No, because \( 6^n = 2^n \times 3^n \), it has no factor of 5, so not divisible by 10, hence cannot end with 0.
6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
7 × 11 × 13 + 13 = 13(7 × 11 + 1) = 13 × 78, product of two numbers greater than 1, so composite.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5040 + 5 = 5045 = 5 × 1009, product of two numbers greater than 1, so composite.
7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
They meet again after LCM(18,12) minutes. 18 = \( 2 \times 3^2 \), 12 = \( 2^2 \times 3 \), LCM = \( 2^2 \times 3^2 = 36 \) minutes.
Exercise 1.2
1. Prove that \( \sqrt{5} \) is irrational.
Assume \( \sqrt{5} = \frac{a}{b} \) where a and b are coprime integers, b ≠ 0. Then \( a^2 = 5b^2 \). So 5 divides a^2, hence 5 divides a (by Theorem 1.2). Let a = 5c, then \( 25c^2 = 5b^2 \), so \( b^2 = 5c^2 \), 5 divides b^2, hence b. Contradiction as a and b coprime. Thus \( \sqrt{5} \) irrational.
2. Prove that \( 3 + 2\sqrt{5} \) is irrational.
Assume \( 3 + 2\sqrt{5} = r \) rational. Then \( 2\sqrt{5} = r - 3 \) rational, so \( \sqrt{5} = \frac{r-3}{2} \) rational, contradiction since \( \sqrt{5} \) irrational.
3. Prove that the following are irrationals:
(i) \( \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \). Assume rational, then \( \sqrt{2} \) = rational × 2 = rational, contradiction.
(ii) \( 7\sqrt{5} \). Assume rational, then \( \sqrt{5} \) = rational / 7 = rational, contradiction.
(iii) \( 6 + \sqrt{2} \). Assume rational, then \( \sqrt{2} \) = rational - 6 = rational, contradiction.
